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AST1100 Lecture Notes
21: Quantum gases
In this lecture we will discuss a quantum phenomenon known as degeneration. A gas which is degenerate has very special properties. The helium
flash was caused by a degenerate stellar core. We will now see that the final
result of stellar evolution is also a star which consists of degenerate gas: a
white dwarf or a neutron star. What gives the degenerate gas its peculiar
properties is its equation of state. The equation of state is an equation relating the pressure of the gas to its density and temperature. We have already
encountered two such equations of state, P = ρkT /(µmH ) for an ideal gas
and P = (1/3)aT 4 for a photon gas (we will now use capital P to denote
pressure in order to distinguish it from the momentum p). We will therefore
start by studying how we can obtain an expression for the pressure of a gas.
1
Pressure
To calculate the pressure of a gas we need to consider the force that particles
in a gas exert on a wall (real or imaginary). Pressure is defined as the force
per area on the wall P = F/A from the particles in the gas. In figure 1 we
see particles within a cylinder of length ∆x. The particles collide with the
wall at the right end of the cylinder. The surface area of the part of the wall
limiting the cylinder on the right end is A. We will assume that the collisions
are elastic. In elastic collisions the total absolute value of the momentum of
the particle is conserved.If we define the x-direction as the direction towards
the wall and the y-direction as the direction along the wall (see again figure
1), then momentum py in y-direction is always conserved since there is no
force working in that direction. For the absolute value of the momentum to
be conserved, the absolute value of the momentum px in the x-direction must
be conserved. This means that the momentum in the x-direction after the
collision must the −px where px is the momentum before the collision. This
means simply that the incoming angle equals the outgoing angle (see figure
2).
Before the collision, a particle has momentum (px , py ). After the collision
1
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∆x
A
Figure 1: Pressure on the wall of area A is the total force exerted from the
particles within the cylinder divided by the area.
px
py
p
θ
θ
− px
p
py
Figure 2: An elastic collision: the absolute value of the momentum is conserved.
2
the particle has momentum (−px , py ). The total change in momentum is just
2px . The force exerted on the wall during a time interval ∆t is according to
Newton’s second law
dp
∆p
2px
f=
≈
=
.
(1)
dt
∆t
∆t
We will now consider a time interval ∆t such that all the particles within
the cylinder with velocity |vx | and momentum |px | has collided with the wall
within time ∆t. The time it takes a particle with x-velocity vx at a distance
∆x from the wall to hit the wall is ∆t′ = ∆x/vx . Within the time ∆t′ , all
the particles in the cylinder traveling in the direction towards the wall (with
velocity vx ) have collided with the wall. But only half of the particles travel
in the direction towards the wall. The other half travels in the opposite
direction and had therefore already hit the wall within a time interval ∆t′
earlier. So within the time
∆t = 2∆t′ =
2∆x
vx
every single particle in the cylinder with velocity |vx | has collided with the
wall. Inserting this in equation 1 we find that the force exerted by any single
particle of velocity |vx | and momentum |px | in the cylinder within the time
∆t is
2px
vx px
f=
=
.
∆t
∆x
The total velocity is given by v 2 = vx2 + vy2 + vz2 , but on average the velocity
components are equally distributed among all three dimensions
such that
√
2
2
2
2
2
< vx >=< vy >=< vz > giving that v = 3vx or vx = v/ 3. Exactly the
√
same argument holds for the momentum giving px = p/ 3. We thus have
1
px vx = pv.
3
such that
vp
.
3∆x
We have a distribution function n(p) which gives us the number density of
particles in the gas with momentum p such that n(p)dp is the number of
particles with momentum between p and p + dp. We have already seen at
least one example of such a distribution function: The Maxwell-Boltzmann
distribution function for an ideal gas. We used this for instance to find the
f=
3
width of a spectral line as well as to find the number of particles with a
certain energy in a stellar core when calculating nuclear reaction rates. We
will here assume a general distribution function n(p). Since n(p) is a number
density, i.e. number per unit volume, we can write the total particles in the
cylinder N(p) with momentum p as the density times the volume A∆x of
the cylinder
N(p) = n(p)A∆x.
The total force exerted on the wall of the cylinder by particles of momentum
p is then
pv
1
dF =
N(p)dp = p v n(p) A dp,
3∆x
3
or in terms of the pressure exerted by these particles
dP =
dF
1
= p v n(p) dp.
A
3
(2)
We obtain the total pressure by integrating
P =
1
3
Z
0
∞
p v n(p) dp
(3)
which is the pressure integral. Given the distribution function n(p) (for instance the Maxwell-Boltzmann distribution) and an expression relating v
and momentum p (for instance v = p/m for non-relativistic particles) we can
integrate this equation to obtain the pressure in the gas.
2
Distribution functions
A statistical distribution function n(p) describes how the momenta are distributed between the particles in a gas. It tells us the number density of
particles having a specific momentum p. The density of particles with momentum between p and p + dp is given by n(p)dp. By making substitutions
(for instance p = mv), we can obtain the velocity distribution function n(v)
which we used to obtain the width of a spectral line in the lectures on electromagnetic radiation. Or by making the substitution E = p2 /(2m) we can
obtain the distribution function n(E) giving the number density of particles
having a certain energy E. We used the latter in the lecture on nuclear
reactions.
4
pz
p
py
px
Figure 3: Momentum space: All particles with |p| within [p, p + ∆p] are
located on the thin shell of thickness ∆p at radius |p|.
The Maxwell-Boltzmann distribution function for an ideal gas is
1
n(~p) = n
2πmkT
3/2
e−p
2 /(2mkT )
,
where n is the total number of particles per volume. This is the density n(~p)
of particles with momentum ~p. Above we needed the number density n(p)
of particles with an absolute value of the momentum p. Thus, we need to
integrate over all possible angles of the vector ~p. We can imagine that we
have a momentum space, i.e. a three dimensional space with axes px , py and
pz (see figure 3). All possible momentum vectors p~ are vectors pointing to
a coordinate (px , py , pz ) in this momentum space. All particles which have
an absolute value p of their momentum p~ are located on a spherical shell at
distance p from the origin in this momentum space. Thus we may imagine a
particle to have a position in the six dimensional position-momentum space
(~x, ~p). All particles have a position in real space (x, y, z) and a position in
momentum space (px , py , pz ). All particles with momentum between p and
p + dp are located on a thin shell of thickness dp at a distance p from the
origin. The total volume of this shell is 4πp2 dp. Thus, to obtain the total
number of particles within this momentum range, we need to multiply the
5
distribution with the momentum space volume 4πp2 dp,
3/2
1
2
e−p /(2mkT ) 4πp2 dp.
2πmkT
This is the distribution function for absolute momenta p. Note that whereas
n(~p) has dimensions number density per real volume per volume in momentum space, n(p)dp has dimensions number density per real volume. The latter
follows from the fact that we have simply multiplied n(~p) with a volume in
momentum space (4πp2 dp) to obtain n(p)dp.
We have also, without knowing it, encountered another distribution function in this course. The Planck distribution. The Planck distribution is the
number density of photons within a given frequency range
n(p)dp = n
2hν 3
1
.
2
hν/(kT
) −1
c e
When you have taken courses in quantum mechanics and thermodynamics you will deduce two more general distribution functions. When taking
quantum mechanical effects into account it can be shown that the distribution function for fermions (fermions were particle with half integer quantum
spin like the electron, proton or neutron) and bosons (bosons were particles
with integer quantum spin like the photon) can be written generally as
B(ν) =
n(E) =
g(E)
e(E−µC )/(kT )
±1
,
where µC is the chemical potential and g(E) is the density of states which we
will come back to later. Here the minus sign is for bosons and the plus sign
for fermions. In the limit of low densities it turns out (we will not show it
here) that the exponential part dominates and the distribution function becomes equal for fermions and bosons. In this case the chemical potential has
such a form that we get back the Maxwell-Boltzmann distribution function
(compare with the above expression). Note that the expression for bosons
resembles the Planck function: the Planck function can be derived from the
distribution function for bosons (you will do this in later courses).
3
Degenerate gases
In the core of stars, the fermions, i.e. the electrons, is the dominating species.
Therefore we will here study the distribution function for fermions and use
6
the + sign in the above equation. We look at an approximation of the
distribution function for an electron gas at low temperature. Of course, the
temperature in the core of a star is not particularly low, but we will later show
that the same approximation and results are valid even for high temperatures
provided we are in the high density limit. In the low temperature limit it can
be shown that the chemical potential µC equals the so-called Fermi energy
EF . We will later find an expression for the Fermi energy, but first we will
consider the distribution function
n(E) =
g(E)
e(E−EF )/(kT )
+1
(4)
where
2me 3/2 1/2
E ,
(5)
h2
where me is the electron mass. The number of electrons per volume with an
energy between E and E + dE in a gas with temperature T and Fermi energy
EF is now given by n(E)dE. The energy E of the electron may be larger or
smaller than the Fermi energy EF . We will now measure the energy of the
electron in units of the Fermi energy. We define x = E/EF such that x < 1
when the energy is less than the Fermi energy and x > 1 when the energy is
larger than the Fermi energy. The distribution function as a function of x,
the energy in units of the Fermi energy, can thus be written
g(E) = 4π
n(x) =
g(x)
e(x−1)EF /(kT )
+1
.
In the low temperature limit, T → 0, the factor EF /(kT ) is a very large
quantity. The energy x defines whether the number in the exponential is a
large positive or a large negative quantity. If x > 1, i.e. that the energy is
larger than the Fermi energy, then the number in the exponential is a large
positive number and n(x) → 0. For x < 1, i.e. the energy is less than the
Fermi energy, the number in the exponential is a large negative number. Thus
the exponential goes to zero and n(x) → g(x). So for very low temperatures,
there is a sharp limit at x = 1. For E < EF we find n(x) = g(x) whereas for
E > EF we find n(x) = 0. In figure 4 we show n(x)/g(x) for lower and lower
temperatures.
The physical meaning of this is that for very low temperatures, all the
electrons have energies up to the Fermi energy whereas no electrons have energies larger than the Fermi energy. The Fermi energy is a low temperature
7
Figure 4: The number of electrons n(E) divided by g(E) for different energies
E. The solid line is for a gas at temperature T = 10K, the dotted line for
a gas at temperature T = 2K and the dashed line for T = 0.1K. When the
temperature approaches zero, there are less and less electrons with energy
larger than the Fermi energy EF .
8
energy limit for the electrons. Even if we cool an electron gas down to zero
temperature, there will still be electrons having energies all the way up to
the Fermi energy. But if the temperature is zero, why don’t all electrons have
an energy close to zero? Why don’t all electrons go and occupy the lowest
possible energy state allowed by quantum mechanics (in quantum mechanics, a particle cannot have zero energy)? The reason for this is hidden in
quantum physics: at low temperatures the gas of electrons start to behave
like a quantum gas, a gas where quantum mechanical effects are important.
The quantum mechanical effect which we see on play here is the Pauli exclusion principle: Two fermions cannot occupy the same energy state. To
understand this principle we need to dig even deeper into the quantum theory. According to quantum mechanics momentum is quantized. This means
that a particle cannot have an arbitrary momentum. The momentum in any
direction can be written as
px = Nx · p0 ,
where Nx is an integer quantum number and p0 is the lowest possible momentum. Thus, an electron can only have x-momenta p0 , 2p0 , 3p0 etc. No
values in between are allowed. So the total momentum of an electron (or any
particle) can be written
p2 = p2x + p2y + p2z = p20 (Nx2 + Ny2 + Nz2 ) ≡ p20 N 2 ,
where (Nx , Ny , Nz ) are the three quantum numbers defining the state of the
electron. According to the Pauli exclusion principle only one electron can
occupy the quantum state (Nx , Ny , Nz ). No other electrons can have exactly the same combination of quantum numbers. We go back to the above
image of a momentum space where a particle has a position (px , py , pz ) in
a three dimensional momentum space in addition to a position in normal
space. We can now write this position in terms of quantum numbers as
(px , py , pz ) = p0 (Nx , Ny , Nz ). Since only one electron can have a given momentum p0 (Nx , Ny , Nz ), one could imagine the momentum space filled with
boxes of volume p0 × p0 × p0 . Only one electron fits into each box. We remember that all electrons with momentum lower than a given momentum p
is within a sphere with radius p in this momentum space. All electrons with
a higher momentum p are outside of this sphere. But inside the sphere of
radius p, there is only room for 4/3πN 3 boxes of size p30 (total volume of the
momentum space sphere (4/3)πp3 = (4/3)πp30N 3 divided by volume of box
p30 ). If all these boxes are filled, no more electron may settle on a position
9
inside this sphere, it has to remain outside of the sphere. When you lower
the temperature of an electron gas, the electrons loose momentum and start
to occupy the lowest possible momentum states, i.e. they all start to move
towards the origin (0, 0, 0) in momentum space. But when all start to move
towards the origin in momentum space (see again figure 3), all the boxes
around the origin are soon occupied, so the electrons need to remain with
higher momenta at larger distances p from the origin. But if they need to
remain with larger momenta, this means that they also have larger energy:
The same argument therefore applies to energy. The energy states of the
electrons are quantized so not all electrons may occupy the lowest energy
state. For this reason we see that the distribution function for electrons at
low temperatures is a step function: All electrons try to occupy the lowest
possible energy state. The lowest energy states are filled up to the Fermi
energy. If we call pF the Fermi momentum, the momentum corresponding
to the Fermi energy we can imagine that all electrons start to gather around
the origin in momentum space out to the radius pF . All electrons are packed
together inside a sphere of radius pF in momentum space. When you add
more electrons to the gas, i.e. the density of electrons increases, the sphere
in momentum space inside which all the electrons are packed also needs to
expand and the Fermi momentum pF increases. Thus the Fermi momentum
and the Fermi energy are functions of the electron density ne .
Having learned that for very low temperatures, the electrons are packed
together in momentum space in a sphere of radius pF we can find the total
number density (per real space volume) ne of electrons in the gas by summing
up all the boxes of size p30 inside this sphere. We know that all these boxes
are occupied by one electron and that no electrons are outside this sphere.
First we need to know the fermion distribution function n(p) in terms of
momentum rather than in terms of energy which we used above. The fermion
distribution function for momentum can be written
n(~p) =
1
(p2 −p2F )/(2mkT )
e
2
.
+ 1 h3
This is the number density per volume in real space per volume in momentum
space. Considering again the low temperature case, we see, using the same
arguments as before, that n(~p) → 0 for p > pF and n(~p) → 2/h3 for p < pF .
Thus n(~p) is a constant for p < pF and zero for p > pF . In order to obtain
the number density of electrons per real space volume we need to integrate
10
this expression over the momentum space volume. So
ne =
Z
∞
0
n(~p)4πp2 dp =
Z
pF
0
2
8π
4πp2 dp = 3 p2F
3
h
3h
where we integrate over the sphere in momentum space in shells of thickness
dp out to the Fermi momentum pF where n(~p) is a constant (2/h3 ) for p < pF
and is zero for p > pF . We use this result to obtain an expression for the
Fermi momentum
!1/3
3h3 ne
.
(6)
pF =
8π
Using the non-relativistic expression for energy we can now find the Fermi
energy expressed in terms of the electron number density ne
p2F
h2
EF =
=
2me
8me
3ne
π
2/3
.
(7)
As we anticipated, the Fermi energy depends on the density of electrons. The
higher the density, the larger the Fermi energy and the Fermi momentum in
order to have space for all the electrons within the sphere of radius pF . A
gas where all particles are packed within this sphere so that the particles
are fighting for a box in momentum space among the lowest energy states
is called a degenerate gas. A partially degenerate gas is a gas where there
are still a few vacant boxes among the lowest energy states such that some
particles have energies larger than the Fermi energy. We now need to find a
criterion for when a gas is degenerate.
When the temperature of a gas is high and the density low, the distribution function is the Maxwell-Boltzmann distribution function. We have
previously learned that for a gas following the Maxwell-Boltzmann distribution function, the mean energy per particle is < E >= (3/2)kT . The gas
starts to become degenerate when most of the particles have energies below
the Fermi energy. The gas therefore starts to be degenerate when the mean
energy of the particles go below the Fermi energy. For an electron gas we
thus have the criterion
3
h2
kT < EF =
2
8me
or
T
2/3
ne
<
h2
12me k
11
3ne
π
2/3
2/3
3
π
.
,
(8)
As discussed above, this criterion is satisfied for very low temperatures, but it
is also satisfied for very high densities. In the exercises you will estimate what
kind of densities are needed in the stellar cores for the core to be degenerate.
4
The pressure of a degenerate electron gas
When the density of electrons in the stellar core becomes high enough, most
electrons have energies below the Fermi energy and the above criterion for
degeneracy is satisfied. The core is electron degenerate. Now we will study
the properties of a degenerate gas. The equation of state, the equation for
the pressure as a function of density and temperature, is one of the most
important properties describing how a gas behaves.
In order to find the pressure, we need to evaluate the pressure integral
(equation 3) for the degenerate gas. First we need the density n(p)dp of
electrons per volume with momentum p in the interval [p, p + dp]. By now
we have learned that n(~p)4πp2 dp = n(p)dp such that for p < pF we have
n(p)dp = (2/h3 )4πp2 dp and for p > pF we have n(p) = 0.
P =
1
3
Z
0
∞
p v n(p) dp =
1
3
Z
pF
0
p2 2
8π 1 5
4πp2 dp =
p .
3
me h
3me h3 5 F
Inserting the expression for the Fermi momentum (equation 6), we find
P =
2/3
3
π
h2 5/3
n
20me e
(9)
We see that the pressure of a degenerate gas does not depend on the temperature. If the temperature increases or decreases, the pressure does not
change! This is very different from a normal gas. It means that the degenerate stellar core will not expand or contract as the temperature changes. The
only exception being when the temperature increases so much that the condition (8) for degeneracy is no longer valid and the degeneracy is broken. In
this case, the electrons have gained so much energy that they are not packed
in the sphere of the lowest momentum states in momentum space. The gas
is no longer degenerate and a normal equation of state which depends on the
temperature needs to be used.
We have deduced the pressure of a degenerate gas using the non-relativistic
expressions for energy. The temperature in the stellar cores are often so high
12
that the velocities of the particles are relativistic. Repeating the above deductions using the relativistic expression, we would obtain
1/3
hc 3
P =
8 π
5
n4/3
e .
(10)
Summary
We have seen that if we compress a gas of fermions sufficiently, so that
the degneracy condition (equation 8) is fulfilled, the fermions are packed
together inside a sphere of radius pF in momentum space. All the lowest
energy states of the fermions are occupied up to the Fermi energy EF . This
typically happens when the temperature is very low so that the fermions fall
down to the lowest possible energy states in momentum space. It might also
happen for high temperatures if the density is high enough: In this case there
are so many fermions present within a volume so all fermion states up to EF
are occupied even if the temperature is not particularly low.
A degenerate fermion gas has a degeneracy pressure which is independent
of the temperature of the gas given by equation (9) for a non-relativistic gas
(the particles have non-relativistic velocities) and by equation (10) for a
relativistic gas. This pressure originates from the resistance against being
squeezed further together in real and momentum space and only depends
on the density of the gas. We obtained the expression for the pressure by
inserting the distribution function for a degenerate gas in the pressure integral (equation 3). The distribution function for a degenerate gas took on a
particular form: It is a step function being constant for energies below the
Fermi energy and zero above. This was simply a consequence of the Pauli
exclusion principle, two energy states cannot be occupied at the same time.
When the quantum states of lowest energy are occupied, the fermions need
to occupy states of higher energy. For a completely degenerate gas, the Fermi
energy EF gives within which energy there is room for all fermions at a given
density.
If the temperature increases sufficiently, the fermions gain enough energy
to occupy states well outside the sphere of radius pF in momentum space.
Then there will be vacant low energy states, the condition of degeneracy is
no longer fulfilled and the gas has become non-degenerate following a normal
temperature-dependent equation of state.
13
6
Problems
Problem 1 In the text we used the pressure integral to find the pressure of
a degenerate electron gas. Study the derivation carefully and make sure you
understand every step before embarking on this exercise.
Now we want to find the pressure in a gas which follows a MaxwellBoltzmann distribution function. Find the expression for the Maxwell-Boltzmann
distribution function and use this in the pressure integral. Assume nonrelativistic velocities. Remember also that the distribution function n(p)
used in the pressure integral needs to be normalized such that
Z
0
∞
n(p)dp = n,
where n is the number density of particles per real space volume. You now
have all the information you need to find the pressure of a gas following
Maxwell-Boltzmann statistics so your task is simply: find P as a function of
n and T . These integrals might be useful:
√
Z ∞
3 π
3/2 −x
x e dx =
0
√4
Z ∞
π
x1/2 e−x dx =
2
0
The answer you find for P should be familiar to you.
Problem 2 In the text we found the condition for a gas to be degenerate
in terms of the temperature T of the gas and the number density ne of
electrons. We will now try to rewrite this expression into a condition on the
mass density ρ of the gas.
1. Assume that the gas is neutral, i.e. that there is an equal number of
protons and electrons. Show that this gives
ne =
Zρ
,
AmH
where Z is the average number of protons per nucleus, A is the average
number of nucleons per nucleus, mH is the hydrogen mass and ρ is the
total mass density.
14
2. Find the expression for the condition for degeneracy in terms of the
total mass density ρ instead of ne .
3. Find the minimum density a gas with temperature T = 109 K must
have in order to be degenerate. A typical atom in the gas has the same
number of protons and neutrons.
4. If you compress the Sun into a sphere with radius R and uniform density until it becomes degenerate, what would be the radius R of the
degenerate compressed Sun? This is basically what will happen at the
end of the Sun’s life time. Gravitation will compress it until it becomes
a degenerate white dwarf star. A white dwarf star typically has a radius similar to the radius of the Earth. Does this fit well with your
result?
5. What about Earth? To which radius would you need to press the Earth
in order for it to become degenerate?
Problem 3 The number density per real space volume per momentum
space volume of particles with momentum p~ is given by n(~p) found in the text.
In order to find the number density per real space volume of particles with
absolute momentum p we multiplied n(~p) with an infinitely small volume
element 4πp2 dp and obtained n(p)dp. Go back to the text and make sure
that you understand this transition.
1. Now we will try to find the number density per real space volume of
particles with energy E using the non-relativistic formula for energy
E = p2 /2m. Start with n(p)dp, make the substitution and show that
you arrive at equation 4 with g(E) looking like equation in 5.
2. In the exercises in the previous lecture, we found that the mean kinetic
energy of a particle in an ideal gas is (3/2)kT . Now we will try to find
the mean kinetic energy in a degenerate gas. First of all, repeat what
you did in the exercise in the previous lecture. Now you will repeat the
same procedure, but use n(E) and E directly,
< E >=
Z
0
∞
P (E)EdE.
You will need to find out how P (E) looks like. The answer is
3
< E >= EF .
5
15